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ReverseNumber.java
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65 lines (56 loc) · 1.79 KB
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/**
* Created by devpriyadave on 2/15/18.
*
* Given a 32-bit signed integer, reverse digits of an integer.
*
* Input: 123456
Output: 321
Input: -123
Output: -321
Input: 120
Output: 21
Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
123 % 10 = 3
123 / 10 =12
12 % 10 = 2
*/
public class ReverseNumber {
public int myReverse(int x) {
Integer number = new Integer(x);
boolean isNegative = false;
if(x <= Integer.MIN_VALUE) {
return 0;
}
if(number < 0) {
System.out.println(number);
number = -number;
isNegative = true;
System.out.println(number);
}
char charArray[] =number.toString().toCharArray();
int size = charArray.length;
for(int start=0, end = size-1; start < size/2; start++, end--){
char temp = charArray[start];
charArray[start] = charArray[end];
charArray[end] = temp;
}
String num = new String(charArray);
if(Long.parseLong(num) > Integer.MAX_VALUE) {
return 0;
} else {
return isNegative ? (-1)*Integer.parseInt(num) : Integer.parseInt(num);
}
}
public int reverse(int x) {
long result = 0;
for(; x != 0; x /= 10){
result = result * 10 + x % 10;
}
return result > Integer.MAX_VALUE || result < Integer.MIN_VALUE ? 0 : (int)result;
}
public static void main(String args[]){
ReverseNumber reverseNumber = new ReverseNumber();
int num = reverseNumber.myReverse(-2147483648);
System.out.println(num);
}
}